Lesson I1

I1 Locating roots by change of sign Quiz: AQA Maths, Unit 9

20 questions

Written by Revision Ninja · Last reviewed 11 October 2026

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Lesson I1, Locating roots by change of sign: 20 multiple choice questions for the AQA Maths (7357), Unit 9: Numerical methods, written with Revision Ninja.

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The 20 questions

  1. If f is continuous on [a, b] and f(a) f(b) < 0, what does this guarantee?

    • f(a) = f(b)
    • There is at least one root of f(x) = 0 in (a, b)
    • There is exactly one root in (a, b)
    • There is no root in (a, b)

  2. Why can change of sign fail to detect a double root?

    • f does not change sign at a double root, because the curve touches the axis
    • The function is discontinuous everywhere
    • The derivative is zero everywhere
    • The root is complex

  3. What condition must f satisfy for a change of sign to locate a root reliably?

    • f must be continuous on the interval
    • f must be linear
    • f must be an even function
    • f must have a positive gradient everywhere

  4. If f(a) and f(b) have the same sign, what can be concluded about roots in (a, b)?

    • There are no roots in (a, b)
    • There is an odd number of roots in (a, b)
    • No conclusion can be drawn, since there may be none or an even number of roots
    • There is exactly one root in (a, b)

  5. Which interval must contain a root of x^3 - x - 1 = 0 by change of sign?

    • (-1, 0)
    • (2, 3)
    • (0, 1)
    • (1, 2)

  6. Which of these intervals shows a change of sign for f(x) = x^2 - 3?

    • (0, 1)
    • (3, 4)
    • (4, 5)
    • (1, 2)

  7. Which interval must contain a root of x^3 - 2x - 5 = 0?

    • (2, 3)
    • (-1, 0)
    • (0, 1)
    • (3, 4)

  8. Which interval must contain a root of cos x - x = 0?

    • (0, 1)
    • (2, 3)
    • (-1, 0)
    • (1, 2)

  9. Which interval must contain a root of ln x + x - 2 = 0?

    • (0, 1)
    • (3, 4)
    • (1, 2)
    • (2, 3)

  10. Which interval must contain a root of x^3 - 4x + 1 = 0?

    • (-1, 0)
    • (3, 4)
    • (0, 1)
    • (1, 1.5)

  11. Which intervals contain the positive root of x^2 - 2x - 1 = 0?

    • (2, 3)
    • (3, 4)
    • (4, 5)
    • (1, 2)

  12. Why does change of sign fail to guarantee a root for f(x) = 1/x on [-1, 1]?

    • The interval is too short
    • f(1) is not positive
    • f has no derivative anywhere
    • f is discontinuous at x = 0, so the sign change comes from a pole rather than a root

  13. For f(x) = x^3 - 2 on [1, 2], what is the new interval after one bisection step?

    • [0, 1]
    • [1.5, 2]
    • [1, 1.5]
    • [1, 2.5]

  14. The equation 1/x - 2 = 0 has a root in [0.25, 1]. What is the root?

    • x = 2
    • x = 1
    • x = 0.25
    • x = 0.5

  15. Which interval must contain a root of x^3 - 7 = 0?

    • (0, 1)
    • (-2, -1)
    • (2, 3)
    • (1, 2)

  16. Which interval must contain a root of x^2 - 2x - 1 = 0 for the negative root, using f(-1) = 2 and f(0) = -1?

    • (0, 1)
    • (-1, 0)
    • (-2, -1)
    • (1, 2)

  17. For f(x) = x^3 - 2x - 1, which interval must contain a root?

    • (-3, -2)
    • (1, 2)
    • (0, 1)
    • (2, 3)

  18. Which interval must contain a root of x^3 + x - 3 = 0?

    • (-2, -1)
    • (1, 2)
    • (2, 3)
    • (0, 1)

  19. A student finds f(a) < 0 and f(b) > 0 for a continuous f on [a, b]. What should the student conclude?

    • The equation f(x) = 0 has a root in (a, b)
    • The function is linear on (a, b)
    • The function has a maximum in (a, b)
    • The root is exactly at the midpoint

  20. For f(x) = x^2 - 4x + 4, why can change of sign fail to locate its root at x = 2?

    • f is discontinuous at x = 2
    • f is zero only at x = 0
    • f has a double root at 2, so f does not change sign there
    • f has no real root at all

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